knitr::opts_chunk$set(comment = NA)
library(janitor); library(naniar)
library(bestglm) ## best subsets search using logistic regression
library(broom)
library(car) ## special plot for logistic regression diagnostics
library(caret) ## for confusion matrices
library(cobalt) ## new today: to split factor into indicator variables
library(cutpointr) ## new today: optimizing cutpoints
library(glue) ## combining R code results and text in labels
library(gt)
library(readxl) ## read in data from an Excel file
library(tableone) ## new today: produce a "simple" Table 1
library(rms)
library(easystats); library(tidyverse)
theme_set(theme_bw()) 432 Class 08
Thomas E. Love, Ph.D.
2025-02-06
Today’s Agenda
- The Bechdel-Wallace Test and the Favorite Movies Data
- Data Cleanup: Working with Complete Cases
- Logistic Regression with
glm()and withlrm()- Fitting Five Models and Comparing their Fits
- Evaluating a Logistic Model in detail
- Determining an optimal decision rule cutpoint
- Evaluating Logistic Regression Assumptions
- Making Predictions and Building Interval Estimates
Today’s R Setup
“Favorite Movies” Data
Ingest Data from an Excel Sheet
[1] 228 80Select Today’s Variables
Variables pulled from mov25_full
| Variable | Description (n = 228 movies) |
|---|---|
mov_id |
Movie ID (meaningless code) |
year |
Year movie was released |
mpa |
Motion Picture Association rating |
reviews |
Number of Critic Reviews on rottentomatoes.com |
ebert |
Star Rating (1-4) on RogerEbert.com |
gen_1 |
Gender (M or F) of first listed star of film (IMDB) |
romance |
1 if Romance is in the movie’s IMDB Genre list (else 0) |
action |
1 if Action is in the movie’s IMDB Genre list (else 0) |
bw_rating |
Score (0-3) on the Bechdel-Wallace test |
movie |
Movie Title |
The Bechdel Test
The Bechdel Test, or Bechdel-Wallace Test was popularized by Alison Bechdel’s comic, in a 1985 strip called The Rule.
The Bechdel-Wallace Test is a simple way to gauge the active presence of female characters in Hollywood films and just how well rounded and complete those roles are1.
Passing the Bechdel-Wallace Test
To pass the test, a movie must have all three of the following.
- at least two (named) women
- who talk to each other
- about something besides a man
mov25_0 <- mov25_0 |>
mutate(bechdel = factor(as.numeric(bw_rating == 3)))
mov25_0 |> tabyl(bechdel, bw_rating) bechdel 0 1 2 3 NA_
0 18 61 15 0 0
1 0 0 0 124 0
<NA> 0 0 0 0 10I use 0 and 1 as the categories for every logistic regression outcome.
Some Data Cleanup
- Drop the films missing the
bechdelinformation. - Create an
agevariable and use it instead ofyear.
mov25_0 <- mov25_0 |>
filter(complete.cases(bechdel)) |>
mutate(age = 2025-year)
mov25_0 |> tabyl(bechdel) |> adorn_pct_formatting() bechdel n percent
0 94 43.1%
1 124 56.9%Again, I always use 0 and 1 for a logistic regression outcome.
More Data Cleanup
- Should we collapse the MPA ratings?
Let’s collapse to the two largest categories, plus “Other”
How should we treat ebert rating?
ebert n percent valid_percent
1.0 4 1.8% 2.0%
1.5 3 1.4% 1.5%
2.0 19 8.7% 9.7%
2.5 20 9.2% 10.2%
3.0 47 21.6% 24.0%
3.5 43 19.7% 21.9%
4.0 60 27.5% 30.6%
NA 22 10.1% -mov25_0$ebert
n missing distinct Info Mean pMedian Gmd
196 22 7 0.945 3.204 3.25 0.815
Value 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Frequency 4 3 19 20 47 43 60
Proportion 0.020 0.015 0.097 0.102 0.240 0.219 0.306
For the frequency table, variable is rounded to the nearest 0What shall we do about missing data?
## using bechdel now instead of bw_rating
mov25_0 <- mov25_0 |>
select(mov_id, bechdel, year, mpa, rt_reviews, ebert, gen_1,
romance, action, bw_rating, movie)
miss_case_table(mov25_0)# A tibble: 2 × 3
n_miss_in_case n_cases pct_cases
<int> <int> <dbl>
1 0 196 89.9
2 1 22 10.1# A tibble: 1 × 3
variable n_miss pct_miss
<chr> <int> <num>
1 ebert 22 10.1Which movies are missing data?
# A tibble: 22 × 3
mov_id ebert movie
<chr> <dbl> <chr>
1 M-001 NA 3 Idiots
2 M-031 NA Castle in the Sky (Tenkû no shiro Rapyuta)
3 M-038 NA Coming To America
4 M-047 NA Dilwale Dulhania Le Jayenge (The Brave Heart Will Take the Brid…
5 M-060 NA A Fistful of Dollars
6 M-087 NA High School Musical 2
7 M-088 NA The Hobbit: An Unexpected Journey
8 M-091 NA Hot Fuzz
9 M-111 NA Kiki's Delivery Service (Majo no takkyûbin)
10 M-130 NA Madea Goes To Jail
# ℹ 12 more rows- Does imputing seem reasonable here?
- Can we assume Missing at Random?
Today, we’ll use complete cases
- Here is my recipe for
mov25from start to finish.
mov25 <- read_xlsx("c08/data/movies_2025-01-28.xlsx", na = c("", "NA")) |>
janitor::clean_names() |>
filter(complete.cases(bw_rating, ebert)) |>
mutate(across(where(is.character), as_factor),
mov_id = as.character(mov_id),
movie = as.character(movie),
age = 2025-year,
mpa3 = fct_lump_n(mpa, n = 2),
bechdel = factor(as.numeric(bw_rating == 3))) |>
select(mov_id, bechdel, age, mpa3, reviews = rt_reviews, ebert,
gen_1, romance, action, movie)
dim(mov25)[1] 196 10mov25 Variable Descriptions
| Variable | Description (n = 196 movies) |
|---|---|
mov_id |
Movie ID (meaningless code) |
bechdel |
1 (“Pass”) or 0 (“Fail”) the Bechdel-Wallace test |
age |
Age of movie (2025 - Year of release) |
mpa3 |
MPA rating (3 levels: PG-13, R, Other) |
reviews |
Number of Critic Reviews on rottentomatoes.com |
ebert |
Star Rating (1-4) on RogerEbert.com |
gen_1 |
Gender (M or F) of first listed star of film (IMDB) |
romance |
1 if Romance is in the movie’s IMDB Genre list (else 0) |
action |
1 if Action is in the movie’s IMDB Genre list (else 0) |
movie |
Movie Title |
data_codebook()
select(mov25, -mov_id, -movie) (196 rows and 8 variables, 8 shown)
ID | Name | Type | Missings | Values | N
---+---------+-------------+----------+-----------+------------
1 | bechdel | categorical | 0 (0.0%) | 0 | 83 (42.3%)
| | | | 1 | 113 (57.7%)
---+---------+-------------+----------+-----------+------------
2 | age | numeric | 0 (0.0%) | [1, 83] | 196
---+---------+-------------+----------+-----------+------------
3 | mpa3 | categorical | 0 (0.0%) | PG-13 | 67 (34.2%)
| | | | R | 61 (31.1%)
| | | | Other | 68 (34.7%)
---+---------+-------------+----------+-----------+------------
4 | reviews | numeric | 0 (0.0%) | [25, 602] | 196
---+---------+-------------+----------+-----------+------------
5 | ebert | numeric | 0 (0.0%) | [1, 4] | 196
---+---------+-------------+----------+-----------+------------
6 | gen_1 | categorical | 0 (0.0%) | M | 149 (76.0%)
| | | | F | 47 (24.0%)
---+---------+-------------+----------+-----------+------------
7 | romance | numeric | 0 (0.0%) | 0 | 151 (77.0%)
| | | | 1 | 45 (23.0%)
---+---------+-------------+----------+-----------+------------
8 | action | numeric | 0 (0.0%) | 0 | 142 (72.4%)
| | | | 1 | 54 (27.6%)
---------------------------------------------------------------The mov25 data listing
# A tibble: 196 × 10
mov_id bechdel age mpa3 reviews ebert gen_1 romance action movie
<chr> <fct> <dbl> <fct> <dbl> <dbl> <fct> <dbl> <dbl> <chr>
1 M-002 1 62 Other 60 4 M 0 0 8 1/2
2 M-003 1 26 PG-13 92 2.5 M 1 0 10 Things I Ha…
3 M-004 0 57 Other 118 4 M 0 0 2001: A Space …
4 M-005 1 16 Other 72 4 F 0 0 About Elly (Da…
5 M-006 1 12 R 168 2.5 M 1 0 About Time
6 M-007 1 46 R 135 4 F 0 0 Alien
7 M-008 1 41 Other 154 4 M 0 0 Amadeus
8 M-009 1 16 PG-13 335 4 M 0 1 Avatar
9 M-010 1 7 PG-13 492 2.5 M 0 1 Avengers: Infi…
10 M-011 1 6 PG-13 556 3 M 0 1 Avengers: Endg…
# ℹ 186 more rows[1] TRUESplitting the sample?
We have 196 films in our mov25 tibble.
- It turns out that a logistic regression model needs about 96 observations just to fit a reasonable intercept term.
- Each additional coefficient we fit requires another 10-20 observations just so that we might validate well.
Here, we want to explore seven predictors (age, mpa3, reviews, ebert, gen_1, romance and action.)
- Does it make sense to split our data into separate training and testing samples?
Four of Dr. Love’s 125 favorite movies
love4 <- tibble(
mov_id = c("L-2", "L-8", "L-63", "L-125"), age = c(63, 37, 22, 30),
bechdel = c(0, 0, 1, 1), gen_1 = c("M", "M", "M", "F"),
mpa3 = c("PG-13", "R", "Other", "Other"), reviews = c(67, 89, 230, 67),
ebert = c(4, 2, 3.5, 3.5), romance = c(0, 0, 1, 1), action = c(0, 1, 0, 0),
movie = c("The Manchurian Candidate", "Die Hard", "Love Actually",
"Sense and Sensibility")) |>
mutate(across(where(is.character), as_factor),
mov_id = as.character(mov_id),
movie = as.character(movie))
love4# A tibble: 4 × 10
mov_id age bechdel gen_1 mpa3 reviews ebert romance action movie
<chr> <dbl> <dbl> <fct> <fct> <dbl> <dbl> <dbl> <dbl> <chr>
1 L-2 63 0 M PG-13 67 4 0 0 The Manchurian …
2 L-8 37 0 M R 89 2 0 1 Die Hard
3 L-63 22 1 M Other 230 3.5 1 0 Love Actually
4 L-125 30 1 F Other 67 3.5 1 0 Sense and Sensi…The Logistic Regression Model
\[ logit(event) = log\left( \frac{Pr(event)}{1 - Pr(event)} \right) = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + ... + \beta_k X_k \]
\[ odds(event) = \frac{Pr(event)}{1 - Pr(event)} \]
\[ Pr(event) = \frac{odds(event)}{odds(event) + 1} \]
\[ Pr(event) = \frac{exp(logit(event))}{1 + exp(logit(event))} \]
Here, our event will be “movie passes the Bechdel-Wallace test” (bechdel = 1)
Guess the associations?
In which direction do you think these variables will be associated with passing the Bechdel-Wallace test?
| Variable | Description (n = 196 movies) |
|---|---|
age |
Age of movie (2025 - Year of release) |
mpa3 |
MPA rating (3 levels: PG-13, R, Other) |
reviews |
Number of Critic Reviews on rottentomatoes.com |
ebert |
Star Rating (1-4) on RogerEbert.com |
gen_1 |
Gender (M or F) of first listed star of film (IMDB) |
romance |
1 if Romance is in the movie’s IMDB Genre list (else 0) |
action |
1 if Action is in the movie’s IMDB Genre list (else 0) |
Table 1?
CreateTableOne(data = mov25,
vars = c("age", "mpa3", "reviews", "ebert", "gen_1", "romance", "action"),
factorVars = c("mpa3", "gen_1", "romance", "action"),
strata = "bechdel") Stratified by bechdel
0 1 p test
n 83 113
age (mean (SD)) 24.42 (15.50) 20.05 (13.13) 0.034
mpa3 (%) 0.387
PG-13 25 (30.1) 42 (37.2)
R 30 (36.1) 31 (27.4)
Other 28 (33.7) 40 (35.4)
reviews (mean (SD)) 183.84 (91.46) 227.20 (135.10) 0.012
ebert (mean (SD)) 3.33 (0.66) 3.11 (0.80) 0.041
gen_1 = F (%) 3 ( 3.6) 44 (38.9) <0.001
romance = 1 (%) 11 (13.3) 34 (30.1) 0.009
action = 1 (%) 29 (34.9) 25 (22.1) 0.068 What Five Models Will We Fit Today?
fit1: two predictors (age,gen_1)fit2: all seven predictors (age,gen_1,mpa3,reviews,ebert,romance,action)fit3: predictor subset with the lowest BIC (frombestglmsearch)fit4: predictor subset with the lowest AIC (bestglm)fit5: add two non-linear terms using our 7 predictors
Model fit1
Model fit1: two predictors
[1] 196AUC: 71.99%# Indices of model performance
AIC | AICc | BIC | Tjur's R2 | RMSE | Sigma | Log_loss | Score_log
------------------------------------------------------------------------------
231.153 | 231.278 | 240.987 | 0.181 | 0.447 | 1.000 | 0.574 | -126.197
AIC | Score_spherical | PCP
---------------------------------
231.153 | 0.005 | 0.600Model fit1 parameters
Parameter | Odds Ratio | SE | 90% CI | z | p
-----------------------------------------------------------------
(Intercept) | 1.33 | 0.41 | [0.81, 2.21] | 0.94 | 0.348
age | 0.98 | 0.01 | [0.96, 1.00] | -1.67 | 0.095
gen 1 [F] | 16.55 | 10.28 | [6.64, 53.82] | 4.52 | < .001or, if you prefer…
# A tibble: 3 × 7
term estimate std.error statistic p.value conf.low conf.high
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 (Intercept) 1.33 0.305 0.938 0.348 0.808 2.21
2 age 0.981 0.0116 -1.67 0.0948 0.962 0.999
3 gen_1F 16.6 0.621 4.52 0.00000619 6.64 53.8 What is the fit1 equation?
(Intercept) age gen_1F
0.28623530 -0.01932756 2.80648551 \[ logit(\mbox{bechdel = 1}) = \\ log\left( \frac{Pr(\mbox{bechdel = 1})}{1 - Pr(\mbox{bechdel = 1})} \right) = \\ 0.2862 -0.0193 (\mbox{age}) + 2.8065 (\mbox{gen_1 = F}) \]
lrm version of fit1
Key Summaries for fit1_lrm include…
- C = 0.720, Nagelkerke \(R^2\) = 0.259, Brier score = 0.200
- See next slide for details.
fit1_lrm summaries
Logistic Regression Model
lrm(formula = bechdel ~ age + gen_1, data = mov25, x = TRUE,
y = TRUE)
Model Likelihood Discrimination Rank Discrim.
Ratio Test Indexes Indexes
Obs 196 LR chi2 41.95 R2 0.259 C 0.720
0 83 d.f. 2 R2(2,196)0.184 Dxy 0.440
1 113 Pr(> chi2) <0.0001 R2(2,143.6)0.243 gamma 0.446
max |deriv| 3e-06 Brier 0.200 tau-a 0.216
Coef S.E. Wald Z Pr(>|Z|)
Intercept 0.2862 0.3051 0.94 0.3482
age -0.0193 0.0116 -1.67 0.0948
gen_1=F 2.8065 0.6209 4.52 <0.0001 ROC Curve Analysis for fit1
Nagelkerke \(R^2\) for fit1
From the lrm() fit (fit1_lrm), we have: Nagelkerke \(R^2\) = 0.259, which isn’t 25.9% of anything, but will be 1 if the fitted model shows as much improvement as possible over the null model.
- This is the version of \(R^2\) that the
validate()function will look at. - The validated version of this is the best way I have to compare models.
Tjur’s \(R^2\) for fit1
From the glm() fit (fit1), and the model_performance() function, we have: Tjur’s \(R^2\) = 0.181.
- Tjur’s \(R^2\) = the difference between the average fitted probability when the outcome is 1 (bechdel Pass) minus the average fitted probability when the outcome is 0 (bechdel Fail, here.)
- It’s less useful than a validated Nagelkerke \(R^2\) if your goal is to compare two models.
The McFadden \(R^2\) for fit1
- The McFadden \(R^2\), which is 1 minus the ratio of (the model deviance over the deviance for the null model.)
[1] 267.1038 225.1527[1] 0.157059The McFadden \(R^2\) (= 0.157 here) and approximates a proportionate reduction in error.
- Also less useful than a validated Nagelkerke \(R^2\) if your goal is to compare two models.
Key fit1 Pseudo-\(R^2\) measures
r2_fit1 <- tibble( name = c("McFadden", "Nagelkerke", "Tjur"),
r_sq = c(as.numeric(r2_mcfadden(fit1)[1]),
as.numeric(r2_nagelkerke(fit1)),
as.numeric(r2_tjur(fit1))))
r2_fit1 |> gt() |> fmt_number(decimals = 4) |>
tab_options(table.font.size = 24) |>
opt_stylize(style = 2, color = "pink")| name | r_sq |
|---|---|
| McFadden | 0.1571 |
| Nagelkerke | 0.2590 |
| Tjur | 0.1813 |
What are the effect sizes in fit1_lrm?
Effects Response : bechdel
Factor Low High Diff. Effect S.E. Lower 0.95 Upper 0.95
age 11 28.25 17.25 -0.33340 0.19955 -0.72451 0.057713
Odds Ratio 11 28.25 17.25 0.71648 NA 0.48456 1.059400
gen_1 - F:M 1 2.00 NA 2.80650 0.62091 1.58950 4.023500
Odds Ratio 1 2.00 NA 16.55200 NA 4.90140 55.894000 Plotting the Effects for fit1_lrm
Prediction Plots for fit1_lrm
Calibration Plot for fit1_lrm
n=196 Mean absolute error=0.018 Mean squared error=0.00046
0.9 Quantile of absolute error=0.036
fit1 Hosmer-Lemeshow test
- This is a somewhat out-of-date method for assessing calibration. Split the data into (here) 5 bins (default is 10) and look for a \(\chi^2\) test result that has a large \(p\) value.
Nomogram for fit1_lrm
Predictions from fit1 for love4 movies
augment(fit1, newdata = love4, type.predict = "response") |>
select(movie, .fitted, bechdel, everything()) |>
gt() |> tab_options(table.font.size = 20) |>
opt_stylize(style = 5, color = "pink")| movie | .fitted | bechdel | mov_id | age | gen_1 | mpa3 | reviews | ebert | romance | action |
|---|---|---|---|---|---|---|---|---|---|---|
| The Manchurian Candidate | 0.2826406 | 0 | L-2 | 63 | M | PG-13 | 67 | 4.0 | 0 | 0 |
| Die Hard | 0.3943927 | 0 | L-8 | 37 | M | R | 89 | 2.0 | 0 | 1 |
| Love Actually | 0.4653130 | 1 | L-63 | 22 | M | Other | 230 | 3.5 | 1 | 0 |
| Sense and Sensibility | 0.9250408 | 1 | L-125 | 30 | F | Other | 67 | 3.5 | 1 | 0 |
CIs around our predictions?
augment(fit1, newdata = love4, type.predict = "link", se_fit = TRUE) |>
mutate(ci_90_low = .fitted - 1.645 * .se.fit,
ci_90_high = .fitted + 1.645 * .se.fit) |>
select(movie, .fitted, .se.fit, ci_90_low, ci_90_high, bechdel, everything())# A tibble: 4 × 14
movie .fitted .se.fit ci_90_low ci_90_high bechdel mov_id age gen_1 mpa3
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <dbl> <fct> <fct>
1 The Man… -0.931 0.501 -1.76 -0.107 0 L-2 63 M PG-13
2 Die Hard -0.429 0.239 -0.822 -0.0361 0 L-8 37 M R
3 Love Ac… -0.139 0.166 -0.412 0.134 1 L-63 22 M Other
4 Sense a… 2.51 0.604 1.52 3.51 1 L-125 30 F Other
# ℹ 4 more variables: reviews <dbl>, ebert <dbl>, romance <dbl>, action <dbl>Converting from Logit to Probability Scale
For The Manchurian Candidate, our predicted logit(bechdel pass) = -0.9314, with 90% CI (-1.7553, -0.1075).
- If logit(bechdel pass) = -0.9314, then odds(bechdel pass) = exp(-0.9314), and pr(bechdel pass) = exp(-0.9314) / (1 + exp(-0.9314)) = 0.3940 / 1.3940 = 0.283
- If logit(bechdel pass) = -1.7553, then odds(bechdel pass) = exp(-1.7553), and pr(bechdel pass) = exp(-1.7553) / (1 + exp(-1.7553)) = 0.1729 / 1.1729 = 0.147
- If logit(bechdel pass) = -0.1075, then odds(bechdel pass) = exp(-0.1075), and pr(bechdel pass) = exp(-0.1075) / (1 + exp(-0.1075)) = 0.8981 / 1.8981 = 0.473
Predicted prob(bechdel pass) = 0.283 with 90% confidence interval (0.147, 0.473) for The Manchurian Candidate using fit1.
Confusion Matrix: Picking a Decision Rule
- We’ll use
cutpointrto select a decision rule which maximizes “Sensitivity” + “Specificity”.
fit1_aug <- augment(fit1, type.predict = "response")
cp1 <- cutpointr(data = fit1_aug, .fitted, bechdel,
pos_class = 1, neg_class = 0,
method = maximize_metric, metric = sum_sens_spec)
cp1 |> select(direction, optimal_cutpoint, method, sum_sens_spec) |>
gt() |> tab_options(table.font.size = 24) |>
opt_stylize(style = 2, color = "pink")| direction | optimal_cutpoint | method | sum_sens_spec |
|---|---|---|---|
| >= | 0.5328563 | maximize_metric | 1.380744 |
Confusion Matrix for fit1
cm1 <- confusionMatrix(data = factor(fit1_aug$.fitted >= cp1$optimal_cutpoint),
reference = factor(fit1_aug$bechdel == 1), positive = "TRUE")
cm1Confusion Matrix and Statistics
Reference
Prediction FALSE TRUE
FALSE 72 55
TRUE 11 58
Accuracy : 0.6633
95% CI : (0.5925, 0.729)
No Information Rate : 0.5765
P-Value [Acc > NIR] : 0.008018
Kappa : 0.3557
Mcnemar's Test P-Value : 1.204e-07
Sensitivity : 0.5133
Specificity : 0.8675
Pos Pred Value : 0.8406
Neg Pred Value : 0.5669
Prevalence : 0.5765
Detection Rate : 0.2959
Detection Prevalence : 0.3520
Balanced Accuracy : 0.6904
'Positive' Class : TRUE
Percentage of Correct Predictions
# Percentage of Correct Predictions from Logistic Regression Model
Full model: 60.02% [54.27% - 65.78%]
Null model: 51.17% [45.30% - 57.04%]
# Likelihood-Ratio-Test
Chi-squared: 41.951
df: 2.000
p-value: 0.000- Herron’s PCP = sum of predicted probabilities where y=1, plus the sum of (1 - predicted probabilities) where y=0, divided by the number of observations.
check_model for fit1 (1/4)
check_model for fit1 (2/4)
check_model for fit1 (3/4)
check_model for fit1 (4/4)
- Extra details for the last three plots…
OK: No outliers detected.
- Based on the following method and threshold: cook (0.5).
- For variable: (Whole model)OK: Simulated residuals appear as uniformly distributed (p = 0.347).Warning: About 86% of the residuals are inside the error bounds (~95% or higher would be good).Analysis of Deviance for fit1
Wald Statistics Response: bechdel
Factor Chi-Square d.f. P
age 2.79 1 0.0948
gen_1 20.43 1 <.0001
TOTAL 23.02 2 <.0001Note
Remember that this result shows sequential tests, and if you change the order of the predictors, the p values will change.
Validating Key Summaries (fit1)
index.orig training test optimism index.corrected n
Dxy 0.4401 0.4350 0.4314 0.0035 0.4366 300
R2 0.2590 0.2659 0.2539 0.0120 0.2469 300
Intercept 0.0000 0.0000 -0.0159 0.0159 -0.0159 300
Slope 1.0000 1.0000 0.9404 0.0596 0.9404 300
Emax 0.0000 0.0000 0.0162 0.0162 0.0162 300
D 0.2089 0.2166 0.2043 0.0124 0.1966 300
U -0.0102 -0.0102 0.0229 -0.0331 0.0229 300
Q 0.2191 0.2268 0.1814 0.0454 0.1737 300
B 0.1998 0.1980 0.2023 -0.0044 0.2042 300
g 1.2422 1.5587 1.1993 0.3593 0.8829 300
gp 0.2164 0.2128 0.2093 0.0035 0.2129 300- C = 0.5 + Dxy, so validated C for
fit1= 0.5 + (0.4366/2) = 0.7183, validated Nagelkerke \(R^2\) = 0.2469, and validated Brier score B = 0.2042
Cross-Validating fit1 AUC
If we wanted to obtain a cross-validated (with 5 folds) C statistic (AUC) from a glm() fit, we can do so with performance_accuracy()…
Model fit2
Model fit2: all seven predictors
fit2 <- glm(bechdel ~ age + gen_1 + mpa3 + reviews +
ebert + romance + action,
data = mov25, family = binomial(link = "logit"))
n_obs(fit2)[1] 196AUC: 79.21%# Indices of model performance
AIC | AICc | BIC | Tjur's R2 | RMSE | Sigma | Log_loss | Score_log
------------------------------------------------------------------------------
227.984 | 228.952 | 257.487 | 0.255 | 0.426 | 1.000 | 0.536 | -135.916
AIC | Score_spherical | PCP
---------------------------------
227.984 | 0.005 | 0.636Model fit2 parameters
Parameter | Odds Ratio | SE | 90% CI | z | p
---------------------------------------------------------------------
(Intercept) | 0.95 | 0.90 | [0.20, 4.53] | -0.06 | 0.955
age | 1.02 | 0.02 | [0.99, 1.05] | 0.97 | 0.332
gen 1 [F] | 13.28 | 8.50 | [5.13, 44.26] | 4.04 | < .001
mpa3 [R] | 0.74 | 0.34 | [0.35, 1.56] | -0.66 | 0.512
mpa3 [Other] | 1.11 | 0.50 | [0.52, 2.35] | 0.23 | 0.821
reviews | 1.01 | 2.35e-03 | [1.00, 1.01] | 3.06 | 0.002
ebert | 0.59 | 0.16 | [0.37, 0.91] | -1.94 | 0.053
romance | 1.69 | 0.79 | [0.79, 3.71] | 1.11 | 0.265
action | 0.46 | 0.20 | [0.23, 0.93] | -1.79 | 0.074 What is the fit2 equation?
(Intercept) age gen_1F mpa3R mpa3Other reviews
-0.053062946 0.016521393 2.586249907 -0.296613775 0.103018693 0.007152862
ebert romance action
-0.528343279 0.522336854 -0.766365711 \[ logit(\mbox{bechdel = 1}) = \\ -0.0531 + 0.0165 (\mbox{age}) + 2.5862 (\mbox{gen_1 = F}) - 0.2966 (\mbox{mpa3 = R}) + \\ 0.1030 (\mbox{mpa3 = Other}) + 0.0072 (\mbox{reviews}) - 0.5283 (\mbox{ebert}) + \\ 0.5223 (\mbox{romance}) - 0.7663 (\mbox{action}) \]
lrm version of fit2
Key Summaries for fit2_lrm include…
- C = 0.792, Nagelkerke \(R^2\) = 0.340, Brier score = 0.181
- See next slide for details.
fit2_lrm summaries
Logistic Regression Model
lrm(formula = bechdel ~ age + gen_1 + mpa3 + reviews + ebert +
romance + action, data = mov25, x = TRUE, y = TRUE)
Model Likelihood Discrimination Rank Discrim.
Ratio Test Indexes Indexes
Obs 196 LR chi2 57.12 R2 0.340 C 0.792
0 83 d.f. 8 R2(8,196)0.222 Dxy 0.584
1 113 Pr(> chi2) <0.0001 R2(8,143.6)0.290 gamma 0.584
max |deriv| 8e-05 Brier 0.181 tau-a 0.287
Coef S.E. Wald Z Pr(>|Z|)
Intercept -0.0531 0.9450 -0.06 0.9552
age 0.0165 0.0170 0.97 0.3316
gen_1=F 2.5862 0.6403 4.04 <0.0001
mpa3=R -0.2966 0.4518 -0.66 0.5115
mpa3=Other 0.1030 0.4550 0.23 0.8209
reviews 0.0072 0.0023 3.06 0.0022
ebert -0.5283 0.2727 -1.94 0.0527
romance 0.5223 0.4689 1.11 0.2653
action -0.7664 0.4290 -1.79 0.0741 ROC Curve Analysis for fit2
Comparing fit1 to fit2 (ROC)
Pseudo-\(R^2\) measures (models so far)
r2_res12 <- tibble(name = c("fit1", "fit2"),
McFadden = c(as.numeric(r2_mcfadden(fit1)[1]),
as.numeric(r2_mcfadden(fit2)[1])),
Nagelkerke = c(as.numeric(r2_nagelkerke(fit1)),
as.numeric(r2_nagelkerke(fit2))),
Tjur = c(as.numeric(r2_tjur(fit1)),
as.numeric(r2_tjur(fit2))))
r2_res12 |> gt() |> fmt_number(decimals = 4) |>
tab_options(table.font.size = 24) |>
opt_stylize(style = 4, color = "green")| name | McFadden | Nagelkerke | Tjur |
|---|---|---|---|
| fit1 | 0.1571 | 0.2590 | 0.1813 |
| fit2 | 0.2138 | 0.3398 | 0.2549 |
Comparing Model Indices from fit1 and fit2
What are the effect sizes in fit2_lrm?
Effects Response : bechdel
Factor Low High Diff. Effect S.E. Lower 0.95 Upper 0.95
age 11.00 28.25 17.25 0.28499 0.29355 -0.29035 0.8603300
Odds Ratio 11.00 28.25 17.25 1.32980 NA 0.74800 2.3640000
reviews 109.75 282.50 172.75 1.23570 0.40358 0.44465 2.0267000
Odds Ratio 109.75 282.50 172.75 3.44060 NA 1.55990 7.5887000
ebert 3.00 4.00 1.00 -0.52834 0.27268 -1.06280 0.0060977
Odds Ratio 3.00 4.00 1.00 0.58958 NA 0.34549 1.0061000
romance 0.00 1.00 1.00 0.52234 0.46894 -0.39677 1.4414000
Odds Ratio 0.00 1.00 1.00 1.68600 NA 0.67249 4.2268000
action 0.00 1.00 1.00 -0.76637 0.42903 -1.60720 0.0745170
Odds Ratio 0.00 1.00 1.00 0.46470 NA 0.20044 1.0774000
gen_1 - F:M 1.00 2.00 NA 2.58620 0.64033 1.33120 3.8413000
Odds Ratio 1.00 2.00 NA 13.28000 NA 3.78570 46.5840000
mpa3 - PG-13:Other 3.00 1.00 NA -0.10302 0.45498 -0.99475 0.7887200
Odds Ratio 3.00 1.00 NA 0.90211 NA 0.36981 2.2006000
mpa3 - R:Other 3.00 2.00 NA -0.39963 0.42825 -1.23900 0.4397200
Odds Ratio 3.00 2.00 NA 0.67057 NA 0.28968 1.5523000Plotting the Effects for fit2_lrm
Prediction Plots for fit2_lrm
Calibration Plot for fit2_lrm
n=196 Mean absolute error=0.026 Mean squared error=0.00112
0.9 Quantile of absolute error=0.053
fit2 Hosmer-Lemeshow test
- Here we’ll revert to the default choice of 10 bins.
Nomogram for fit2_lrm
Predictions from fit2 for love4 movies
augment(fit2, newdata = love4, type.predict = "response") |>
select(movie, .fitted, bechdel, everything()) |>
gt() |> tab_options(table.font.size = 20) |>
opt_stylize(style = 5, color = "pink")| movie | .fitted | bechdel | mov_id | age | gen_1 | mpa3 | reviews | ebert | romance | action |
|---|---|---|---|---|---|---|---|---|---|---|
| The Manchurian Candidate | 0.3438133 | 0 | L-2 | 63 | M | PG-13 | 67 | 4.0 | 0 | 0 |
| Die Hard | 0.2839742 | 0 | L-8 | 37 | M | R | 89 | 2.0 | 0 | 1 |
| Love Actually | 0.6751826 | 1 | L-63 | 22 | M | Other | 230 | 3.5 | 1 | 0 |
| Sense and Sensibility | 0.9075621 | 1 | L-125 | 30 | F | Other | 67 | 3.5 | 1 | 0 |
CIs around our predictions?
augment(fit2, newdata = love4, type.predict = "link", se_fit = TRUE) |>
mutate(ci_90_low = .fitted - 1.645 * .se.fit,
ci_90_high = .fitted + 1.645 * .se.fit) |>
select(movie, .fitted, .se.fit, ci_90_low, ci_90_high, bechdel, everything())# A tibble: 4 × 14
movie .fitted .se.fit ci_90_low ci_90_high bechdel mov_id age gen_1 mpa3
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <dbl> <fct> <fct>
1 The Man… -0.646 0.719 -1.83 0.536 0 L-2 63 M PG-13
2 Die Hard -0.925 0.620 -1.95 0.0957 0 L-8 37 M R
3 Love Ac… 0.732 0.517 -0.118 1.58 1 L-63 22 M Other
4 Sense a… 2.28 0.732 1.08 3.49 1 L-125 30 F Other
# ℹ 4 more variables: reviews <dbl>, ebert <dbl>, romance <dbl>, action <dbl>Converting from Logit to Probability Scale
For Sense and Sensibility, our predicted logit(bechdel pass) = 2.2842, with 90% CI (1.0809, 3.4876).
- If logit(bechdel pass) = 2.2842, then odds(bechdel pass) = exp(2.2842), and pr(bechdel pass) = exp(2.2842) / (1 + exp(2.2842)) = 9.8178 / 10.8178 = 0.908
- If logit(bechdel pass) = 1.0809, then odds(bechdel pass) = exp(1.0809), and pr(bechdel pass) = exp(1.0809) / (1 + exp(1.0809)) = 2.9473 / 3.9473 = 0.747
- If logit(bechdel pass) = 3.4876, then odds(bechdel pass) = exp(3.4876), and pr(bechdel pass) = exp(3.4876) / (1 + exp(3.4876)) = 32.7074 / 33.7074 = 0.970
Predicted prob(bechdel pass) = 0.908 with 90% confidence interval (0.747, 0.970) for Sense and Sensibility using fit2.
Picking a Decision Rule for fit2
- Again, using
cutpointrto select a decision rule which maximizes “Sensitivity” + “Specificity”.
fit2_aug <- augment(fit2, type.predict = "response")
cp2 <- cutpointr(data = fit2_aug, .fitted, bechdel,
pos_class = 1, neg_class = 0,
method = maximize_metric, metric = sum_sens_spec)
cp2 |> select(direction, optimal_cutpoint, method, sum_sens_spec) |>
gt() |> tab_options(table.font.size = 24) |>
opt_stylize(style = 2, color = "pink")| direction | optimal_cutpoint | method | sum_sens_spec |
|---|---|---|---|
| >= | 0.6950129 | maximize_metric | 1.494829 |
Confusion Matrix for fit2
cm2 <- confusionMatrix(data = factor(fit2_aug$.fitted >= cp2$optimal_cutpoint),
reference = factor(fit2_aug$bechdel == 1), positive = "TRUE")
cm2Confusion Matrix and Statistics
Reference
Prediction FALSE TRUE
FALSE 80 53
TRUE 3 60
Accuracy : 0.7143
95% CI : (0.6456, 0.7764)
No Information Rate : 0.5765
P-Value [Acc > NIR] : 4.645e-05
Kappa : 0.4582
Mcnemar's Test P-Value : 5.835e-11
Sensitivity : 0.5310
Specificity : 0.9639
Pos Pred Value : 0.9524
Neg Pred Value : 0.6015
Prevalence : 0.5765
Detection Rate : 0.3061
Detection Prevalence : 0.3214
Balanced Accuracy : 0.7474
'Positive' Class : TRUE
fit2 PCP
Percentage of Correct Predictions (with 0.5 decision rule)
Checking Assumptions in Logistic Regression Models
Linear Regression vs. Logistic Regression
Adapted from https://www.statology.org/assumptions-of-logistic-regression/
In contrast to linear regression, logistic regression does not require:
- A linear relationship between the predictors and the outcome.
- The residuals of the model to be normally distributed.
- The residuals to have constant variance (homoscedasticity/)
Assumptions of Logistic Regression
Adapted from https://www.statology.org/assumptions-of-logistic-regression/
- The outcome variable is binary.
- The observations are independent from each other. (They shouldn’t show a pattern in time or space.)
- There is no severe multicollinearity among the predictors (we use VIF > 5 as an indicator of trouble.)
Assumptions of Logistic Regression
- There are no extreme outliers (Cook’s distance > .5 is what R flags as problematic1.)
- The sample size is sufficiently large (see next few slides.)
- There is a linear relationship between predictors and the logit of the outcome (see the final few slides.)
What does sufficiently large mean?
- Some people like a simple rule like 500 observations overall and 10 events (where an event is the smaller of your two outcome groups) per predictor parameter. See Long’s 1997 book (pdf).
For Project A, we focus on keeping the number of predictors below (4 + (N-100)) / 100) where N is the size of the smaller of your two outcome groups. I wouldn’t use that standard outside of Project A, though.
What does sufficiently large mean?
- Riley et al. in Statistics in Medicine develop an estimation scheme for the needed sample sizes, and motivate it with several examples. It’s pretty complex but it’s a good option.
First Two Assumptions
- The outcome variable is binary.
- OK.
bechdelis either 1 (Pass) or 0 (Fail.)
- OK.
- The observations are independent from each other. (They shouldn’t show a pattern in time or space.)
- The data are cross-sectional. No one film’s results should affect another film’s results, so we’re OK.
Assumption Three
- There is no severe multicollinearity among the predictors (we use VIF > 5 as an indicator of trouble.)
GVIF Df GVIF^(1/(2*Df))
age 2.058811 1 1.434856
gen_1 1.031006 1 1.015385
mpa3 1.390834 2 1.085973
reviews 2.345235 1 1.531416
ebert 1.279741 1 1.131257
romance 1.131410 1 1.063678
action 1.352148 1 1.162819 age gen_1=F mpa3=R mpa3=Other reviews ebert romance
2.058813 1.031005 1.635040 1.657426 2.345235 1.279742 1.131409
action
1.352148 Assumption Four
- There are no extreme outliers (no Cook’s distance > 0.5)
Assumption Five
- The sample size is sufficiently large.
- Recall that we have 113 Pass and 83 Fail movies in
mov25.
Does this seem like enough observations to fit a logistic regression model with 7 predictors (and 8 df) under consideration?
Assumption Six
- There is a linear relationship between predictors and the logit of the outcome.
A Box-Tidwell test is a common strategy to test this assumption, but it doesn’t work for logistic models according to John Fox, inventor of the car package1.
He instead recommends Component + Residual plots (Partial Residual plots) which can be used for linear and generalized linear models.
Interpreting the Partial Residual Plots
- The blue dashed line shows the expected residuals if the relationship between the predictor and response variable (here the log odds of our outcome) was linear.
- The solid pink curve shows a loess smooth of the actual residuals.
If the two lines are meaningfully different, then this is evidence of a nonlinear relationship. One way to fix this issue is to build a transformation on the predictor variables, or consider incorporating some non-linear terms.
Running Partial Residual Plots
check_model for fit2 (1/4)
check_model for fit2 (2/4)
check_model for fit2 (3/4)
check_model for fit2 (4/4)
- Extra details for the last three plots…
OK: No outliers detected.
- Based on the following method and threshold: cook (0.9).
- For variable: (Whole model)OK: Simulated residuals appear as uniformly distributed (p = 0.234).Warning: About 93% of the residuals are inside the error bounds (~95% or higher would be good).Analysis of Deviance for fit2
Wald Statistics Response: bechdel
Factor Chi-Square d.f. P
age 0.94 1 0.3316
gen_1 16.31 1 0.0001
mpa3 0.93 2 0.6271
reviews 9.37 1 0.0022
ebert 3.75 1 0.0527
romance 1.24 1 0.2653
action 3.19 1 0.0741
TOTAL 32.63 8 0.0001Note
Remember that this result shows sequential tests, and if you change the order of the predictors, the p values will change.
Validating Key Summaries (fit2)
index.orig training test optimism index.corrected n
Dxy 0.5842 0.6152 0.5471 0.0681 0.5161 300
R2 0.3398 0.3780 0.3056 0.0724 0.2674 300
Intercept 0.0000 0.0000 -0.0152 0.0152 -0.0152 300
Slope 1.0000 1.0000 0.7964 0.2036 0.7964 300
Emax 0.0000 0.0000 0.0533 0.0533 0.0533 300
D 0.2863 0.3265 0.2531 0.0734 0.2129 300
U -0.0102 -0.0102 0.0283 -0.0385 0.0283 300
Q 0.2965 0.3367 0.2247 0.1119 0.1846 300
B 0.1811 0.1725 0.1903 -0.0178 0.1989 300
g 1.6424 2.0734 1.4809 0.5925 1.0499 300
gp 0.2827 0.2987 0.2668 0.0318 0.2509 300- C = 0.5 + Dxy, so validated C for
fit2= 0.5 + (0.5161/2) = 0.7581, validated Nagelkerke \(R^2\) = 0.2674, and validated Brier score B = 0.1989
Cross-Validating AUC for fit2
Model fit3
Preparing to Search through our predictors
[1] "mov_id" "bechdel" "age" "mpa3_R" "mpa3_Other"
[6] "reviews" "ebert" "gen_1" "romance" "action"
[11] "movie" - We’re building
Xyto contain all of our predictors (with multi-categorical predictors inserted using indicator variables) followed by the outcome.
Predictor Subset with the best AIC
The code below searches the predictors in Xy for the combination that produces the smallest (hence best) AIC (Akaike Information Criterion).
AIC
BICq equivalent for q in (0.726461340395408, 0.861227305047834)
Best Model:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.529989558 0.827647052 0.640357 5.219406e-01
reviews 0.005135746 0.001600168 3.209506 1.329635e-03
ebert -0.465685210 0.250830225 -1.856575 6.337160e-02
gen_1F 2.711777250 0.632437155 4.287821 1.804345e-05
action -0.715810359 0.381228674 -1.877640 6.043038e-02Model fit3: best subsets with AIC
fit3 <- glm(bechdel ~ gen_1 + reviews + ebert + action,
data = mov25, family = binomial(link = "logit"))
n_obs(fit3)[1] 196AUC: 77.92%# Indices of model performance
AIC | AICc | BIC | Tjur's R2 | RMSE | Sigma | Log_loss | Score_log
------------------------------------------------------------------------------
223.684 | 224.000 | 240.075 | 0.240 | 0.429 | 1.000 | 0.545 | -132.860
AIC | Score_spherical | PCP
---------------------------------
223.684 | 0.008 | 0.629Model fit3 parameters
Parameter | Odds Ratio | SE | 90% CI | z | p
--------------------------------------------------------------------
(Intercept) | 1.70 | 1.41 | [0.44, 6.79] | 0.64 | 0.522
gen 1 [F] | 15.06 | 9.52 | [5.91, 49.67] | 4.29 | < .001
reviews | 1.01 | 1.61e-03 | [1.00, 1.01] | 3.21 | 0.001
ebert | 0.63 | 0.16 | [0.41, 0.94] | -1.86 | 0.063
action | 0.49 | 0.19 | [0.26, 0.91] | -1.88 | 0.060 What is the fit3 equation?
(Intercept) gen_1F reviews ebert action
0.529989558 2.711777250 0.005135746 -0.465685210 -0.715810359 \[ logit(\mbox{bechdel = 1}) = \\ 0.5300 + 2.7118 (\mbox{gen_1 = F}) + 0.0051 (\mbox{reviews}) - \\ 0.4657 (\mbox{ebert}) - 0.7158 (\mbox{action}) \]
lrm version of fit3
Key Summaries for fit3_lrm include…
- C = 0.779, Nagelkerke \(R^2\) = 0.321, Brier score = 0.184
- See next slide for details.
fit3_lrm summaries
Logistic Regression Model
lrm(formula = bechdel ~ gen_1 + reviews + ebert + action, data = mov25,
x = TRUE, y = TRUE)
Model Likelihood Discrimination Rank Discrim.
Ratio Test Indexes Indexes
Obs 196 LR chi2 53.42 R2 0.321 C 0.779
0 83 d.f. 4 R2(4,196)0.223 Dxy 0.558
1 113 Pr(> chi2) <0.0001 R2(4,143.6)0.291 gamma 0.558
max |deriv| 5e-05 Brier 0.184 tau-a 0.274
Coef S.E. Wald Z Pr(>|Z|)
Intercept 0.5300 0.8276 0.64 0.5219
gen_1=F 2.7118 0.6324 4.29 <0.0001
reviews 0.0051 0.0016 3.21 0.0013
ebert -0.4657 0.2508 -1.86 0.0634
action -0.7158 0.3812 -1.88 0.0604 ROC Curve Analysis for fit3
ROC curves for models so far
Pseudo-\(R^2\) measures (models so far)
r2_res123 <- tibble(name = c("fit1", "fit2", "fit3"),
McFadden = c(as.numeric(r2_mcfadden(fit1)[1]),
as.numeric(r2_mcfadden(fit2)[1]),
as.numeric(r2_mcfadden(fit3)[1])),
Nagelkerke = c(as.numeric(r2_nagelkerke(fit1)),
as.numeric(r2_nagelkerke(fit2)),
as.numeric(r2_nagelkerke(fit3))),
Tjur = c(as.numeric(r2_tjur(fit1)),
as.numeric(r2_tjur(fit2)),
as.numeric(r2_tjur(fit3))))
r2_res123 |> gt() |> fmt_number(decimals = 4) |>
tab_options(table.font.size = 24) |>
opt_stylize(style = 4, color = "green")Pseudo-\(R^2\) measures (models so far)
| name | McFadden | Nagelkerke | Tjur |
|---|---|---|---|
| fit1 | 0.1571 | 0.2590 | 0.1813 |
| fit2 | 0.2138 | 0.3398 | 0.2549 |
| fit3 | 0.2000 | 0.3206 | 0.2397 |
Comparing Model Indices from our 3 models
What are the effect sizes in fit3_lrm?
Effects Response : bechdel
Factor Low High Diff. Effect S.E. Lower 0.95 Upper 0.95
reviews 109.75 282.5 172.75 0.88720 0.27643 0.34541 1.429000
Odds Ratio 109.75 282.5 172.75 2.42830 NA 1.41260 4.174500
ebert 3.00 4.0 1.00 -0.46569 0.25083 -0.95730 0.025934
Odds Ratio 3.00 4.0 1.00 0.62770 NA 0.38393 1.026300
action 0.00 1.0 1.00 -0.71581 0.38123 -1.46300 0.031385
Odds Ratio 0.00 1.0 1.00 0.48880 NA 0.23154 1.031900
gen_1 - F:M 1.00 2.0 NA 2.71180 0.63245 1.47220 3.951400
Odds Ratio 1.00 2.0 NA 15.05600 NA 4.35880 52.006000 Plotting the Effects for fit3_lrm
Prediction Plots for fit3_lrm
Calibration Plot for fit3_lrm
n=196 Mean absolute error=0.018 Mean squared error=0.00052
0.9 Quantile of absolute error=0.033
fit3 Hosmer-Lemeshow test
Nomogram for fit3_lrm
Predictions from fit3 for love4 movies
augment(fit3, newdata = love4, type.predict = "response") |>
select(movie, .fitted, bechdel, everything()) |>
gt() |> tab_options(table.font.size = 20) |>
opt_stylize(style = 5, color = "pink")| movie | .fitted | bechdel | mov_id | age | gen_1 | mpa3 | reviews | ebert | romance | action |
|---|---|---|---|---|---|---|---|---|---|---|
| The Manchurian Candidate | 0.2711776 | 0 | L-2 | 63 | M | PG-13 | 67 | 4.0 | 0 | 0 |
| Die Hard | 0.3407150 | 0 | L-8 | 37 | M | R | 89 | 2.0 | 0 | 1 |
| Love Actually | 0.5203171 | 1 | L-63 | 22 | M | Other | 230 | 3.5 | 1 | 0 |
| Sense and Sensibility | 0.8760955 | 1 | L-125 | 30 | F | Other | 67 | 3.5 | 1 | 0 |
CIs around our predictions?
augment(fit3, newdata = love4, type.predict = "link", se_fit = TRUE) |>
mutate(ci_90_low = .fitted - 1.645 * .se.fit,
ci_90_high = .fitted + 1.645 * .se.fit) |>
select(movie, .fitted, .se.fit, ci_90_low, ci_90_high, bechdel, everything())# A tibble: 4 × 14
movie .fitted .se.fit ci_90_low ci_90_high bechdel mov_id age gen_1 mpa3
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <dbl> <fct> <fct>
1 The Man… -0.989 0.358 -1.58 -0.399 0 L-2 63 M PG-13
2 Die Hard -0.660 0.461 -1.42 0.0980 0 L-8 37 M R
3 Love Ac… 0.0813 0.215 -0.272 0.435 1 L-63 22 M Other
4 Sense a… 1.96 0.636 0.909 3.00 1 L-125 30 F Other
# ℹ 4 more variables: reviews <dbl>, ebert <dbl>, romance <dbl>, action <dbl>Converting from Logit to Probability Scale
For Love Actually, our predicted logit(bechdel pass) = 0.0813, with 90% CI (-0.2725, 0.4351).
- If logit(bechdel pass) = 0.0813, then odds(bechdel pass) = exp(0.0813), and pr(bechdel pass) = exp(0.0813) / (1 + exp(0.0813)) = 1.0847/2.0847 = 0.520
- If logit(bechdel pass) = -0.2725, then odds(bechdel pass) = exp(-0.2725), and pr(bechdel pass) = exp(-0.2725) / (1 + exp(-0.2725)) = 0.7615/1.7615 = 0.432
- If logit(bechdel pass) = 0.4351, then odds(bechdel pass) = exp(0.4351), and pr(bechdel pass) = exp(0.4351) / (1 + exp(0.4351)) = 1.5451 / 2.5451 = 0.607
Predicted prob(bechdel pass) = 0.520 with 90% confidence interval (0.432, 0.607) for Love Actually using fit3.
Picking a Decision Rule for fit3
- Again, using
cutpointrto select a decision rule which maximizes “Sensitivity” + “Specificity”.
fit3_aug <- augment(fit3, type.predict = "response")
cp3 <- cutpointr(data = fit3_aug, .fitted, bechdel,
pos_class = 1, neg_class = 0,
method = maximize_metric, metric = sum_sens_spec)
cp3 |> select(direction, optimal_cutpoint, method, sum_sens_spec) |>
gt() |> tab_options(table.font.size = 24) |>
opt_stylize(style = 2, color = "pink")| direction | optimal_cutpoint | method | sum_sens_spec |
|---|---|---|---|
| >= | 0.5631739 | maximize_metric | 1.495788 |
Confusion Matrix for fit3
cm3 <- confusionMatrix(data = factor(fit3_aug$.fitted >= cp3$optimal_cutpoint),
reference = factor(fit3_aug$bechdel == 1), positive = "TRUE")
cm3Confusion Matrix and Statistics
Reference
Prediction FALSE TRUE
FALSE 72 42
TRUE 11 71
Accuracy : 0.7296
95% CI : (0.6617, 0.7904)
No Information Rate : 0.5765
P-Value [Acc > NIR] : 6.311e-06
Kappa : 0.4724
Mcnemar's Test P-Value : 3.775e-05
Sensitivity : 0.6283
Specificity : 0.8675
Pos Pred Value : 0.8659
Neg Pred Value : 0.6316
Prevalence : 0.5765
Detection Rate : 0.3622
Detection Prevalence : 0.4184
Balanced Accuracy : 0.7479
'Positive' Class : TRUE
fit3 PCP
Percentage of Correct Predictions (with 0.5 decision rule)
No severe multicollinearity? (fit3)
No Cook’s distance > 0.5? (fit3)
fit3 Partial Residual Plots
check_model for fit3 (1/4)
check_model for fit3 (2/4)
check_model for fit3 (3/4)
check_model for fit3 (4/4)
- Extra details for the last three plots…
OK: No outliers detected.
- Based on the following method and threshold: cook (0.8).
- For variable: (Whole model)OK: Simulated residuals appear as uniformly distributed (p = 0.265).Warning: About 93% of the residuals are inside the error bounds (~95% or higher would be good).Analysis of Deviance for fit3
Wald Statistics Response: bechdel
Factor Chi-Square d.f. P
gen_1 18.38 1 <.0001
reviews 10.30 1 0.0013
ebert 3.45 1 0.0634
action 3.53 1 0.0604
TOTAL 30.70 4 <.0001Note
Remember that this result shows sequential tests, and if you change the order of the predictors, the p values will change.
Validating Key Summaries (fit3)
index.orig training test optimism index.corrected n
Dxy 0.5584 0.5668 0.5423 0.0245 0.5339 300
R2 0.3206 0.3349 0.3047 0.0301 0.2905 300
Intercept 0.0000 0.0000 0.0038 -0.0038 0.0038 300
Slope 1.0000 1.0000 0.9194 0.0806 0.9194 300
Emax 0.0000 0.0000 0.0194 0.0194 0.0194 300
D 0.2674 0.2831 0.2521 0.0309 0.2365 300
U -0.0102 -0.0102 0.0182 -0.0284 0.0182 300
Q 0.2776 0.2933 0.2339 0.0593 0.2183 300
B 0.1845 0.1811 0.1893 -0.0082 0.1927 300
g 1.5573 1.8436 1.4784 0.3652 1.1921 300
gp 0.2715 0.2757 0.2629 0.0128 0.2587 300- C = 0.5 + Dxy, so validated C for
fit3= 0.5 + (0.5339/2) = 0.7670, validated Nagelkerke \(R^2\) = 0.2905, and validated Brier score B = 0.1927
Cross-Validating AUC for fit3
Model fit4
Preparing to Search through our predictors
[1] "mov_id" "bechdel" "age" "mpa3_R" "mpa3_Other"
[6] "reviews" "ebert" "gen_1" "romance" "action"
[11] "movie" Just as before, Xy contains all predictors (with multi-categorical predictors inserted using indicator variables) followed by the outcome.
Predictor Subset with the best BIC
The code below searches the predictors in Xy for the combination that produces the smallest (hence best) BIC (Bayes Information Criterion).
BIC
BICq equivalent for q in (0.227337090988041, 0.726461340395408)
Best Model:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.959364552 0.345824391 -2.774138 5.534821e-03
reviews 0.003885784 0.001452304 2.675600 7.459551e-03
gen_1F 2.926645287 0.626750302 4.669555 3.018522e-06Model fit4: best subsets with BIC
[1] 196AUC: 74.30%# Indices of model performance
AIC | AICc | BIC | Tjur's R2 | RMSE | Sigma | Log_loss | Score_log
------------------------------------------------------------------------------
226.334 | 226.459 | 236.168 | 0.208 | 0.439 | 1.000 | 0.562 | -129.434
AIC | Score_spherical | PCP
---------------------------------
226.334 | 0.007 | 0.613Model fit4 parameters
Parameter | Odds Ratio | SE | 90% CI | z | p
--------------------------------------------------------------------
(Intercept) | 0.38 | 0.13 | [0.21, 0.67] | -2.77 | 0.006
gen 1 [F] | 18.66 | 11.70 | [7.41, 61.16] | 4.67 | < .001
reviews | 1.00 | 1.46e-03 | [1.00, 1.01] | 2.68 | 0.007 What is the fit4 equation?
(Intercept) gen_1F reviews
-0.959364552 2.926645287 0.003885784 \[ logit(\mbox{bechdel = 1}) = \\ -0.9594 + 2.9266 (\mbox{gen_1 = F}) + 0.0039 (\mbox{reviews}) \]
lrm version of fit4
Key Summaries for fit4_lrm include…
- C = 0.743, Nagelkerke \(R^2\) = 0.285, Brier score = 0.192
- See next slide for details.
fit4_lrm summaries
Logistic Regression Model
lrm(formula = bechdel ~ gen_1 + reviews, data = mov25, x = TRUE,
y = TRUE)
Model Likelihood Discrimination Rank Discrim.
Ratio Test Indexes Indexes
Obs 196 LR chi2 46.77 R2 0.285 C 0.743
0 83 d.f. 2 R2(2,196)0.204 Dxy 0.486
1 113 Pr(> chi2) <0.0001 R2(2,143.6)0.268 gamma 0.487
max |deriv| 3e-05 Brier 0.192 tau-a 0.239
Coef S.E. Wald Z Pr(>|Z|)
Intercept -0.9594 0.3458 -2.77 0.0055
gen_1=F 2.9266 0.6268 4.67 <0.0001
reviews 0.0039 0.0015 2.68 0.0075 ROC Curve Analysis for fit4
ROC curves for models so far
Pseudo-\(R^2\) measures (models so far)
r2_res1234 <- tibble(name = c("fit1", "fit2", "fit3", "fit4"),
McFadden = c(as.numeric(r2_mcfadden(fit1)[1]),
as.numeric(r2_mcfadden(fit2)[1]),
as.numeric(r2_mcfadden(fit3)[1]),
as.numeric(r2_mcfadden(fit4)[1])),
Nagelkerke = c(as.numeric(r2_nagelkerke(fit1)),
as.numeric(r2_nagelkerke(fit2)),
as.numeric(r2_nagelkerke(fit3)),
as.numeric(r2_nagelkerke(fit4))),
Tjur = c(as.numeric(r2_tjur(fit1)),
as.numeric(r2_tjur(fit2)),
as.numeric(r2_tjur(fit3)),
as.numeric(r2_tjur(fit4))))
r2_res1234 |> gt() |> fmt_number(decimals = 4) |>
tab_options(table.font.size = 24) |>
opt_stylize(style = 4, color = "green")Pseudo-\(R^2\) measures (models so far)
| name | McFadden | Nagelkerke | Tjur |
|---|---|---|---|
| fit1 | 0.1571 | 0.2590 | 0.1813 |
| fit2 | 0.2138 | 0.3398 | 0.2549 |
| fit3 | 0.2000 | 0.3206 | 0.2397 |
| fit4 | 0.1751 | 0.2853 | 0.2082 |
Comparing Model Indices from our 4 models
What are the effect sizes in fit4_lrm?
Effects Response : bechdel
Factor Low High Diff. Effect S.E. Lower 0.95 Upper 0.95
reviews 109.75 282.5 172.75 0.67127 0.25089 0.17954 1.1630
Odds Ratio 109.75 282.5 172.75 1.95670 NA 1.19670 3.1995
gen_1 - F:M 1.00 2.0 NA 2.92660 0.62676 1.69820 4.1551
Odds Ratio 1.00 2.0 NA 18.66500 NA 5.46420 63.7560 Plotting the Effects for fit4_lrm
Prediction Plots for fit4_lrm
Calibration Plot for fit4_lrm
n=196 Mean absolute error=0.04 Mean squared error=0.00248
0.9 Quantile of absolute error=0.084
fit4 Hosmer-Lemeshow test
Nomogram for fit4_lrm
Predictions from fit4 for love4 movies
augment(fit4, newdata = love4, type.predict = "response") |>
select(movie, .fitted, bechdel, everything()) |>
gt() |> tab_options(table.font.size = 20) |>
opt_stylize(style = 5, color = "pink")| movie | .fitted | bechdel | mov_id | age | gen_1 | mpa3 | reviews | ebert | romance | action |
|---|---|---|---|---|---|---|---|---|---|---|
| The Manchurian Candidate | 0.3320302 | 0 | L-2 | 63 | M | PG-13 | 67 | 4.0 | 0 | 0 |
| Die Hard | 0.3512544 | 0 | L-8 | 37 | M | R | 89 | 2.0 | 0 | 1 |
| Love Actually | 0.4835973 | 1 | L-63 | 22 | M | Other | 230 | 3.5 | 1 | 0 |
| Sense and Sensibility | 0.9027032 | 1 | L-125 | 30 | F | Other | 67 | 3.5 | 1 | 0 |
CIs around our predictions?
augment(fit4, newdata = love4, type.predict = "link", se_fit = TRUE) |>
mutate(ci_90_low = .fitted - 1.645 * .se.fit,
ci_90_high = .fitted + 1.645 * .se.fit) |>
select(movie, .fitted, .se.fit, ci_90_low, ci_90_high, bechdel, everything())# A tibble: 4 × 14
movie .fitted .se.fit ci_90_low ci_90_high bechdel mov_id age gen_1 mpa3
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <dbl> <fct> <fct>
1 The Man… -0.699 0.265 -1.14 -0.263 0 L-2 63 M PG-13
2 Die Hard -0.614 0.241 -1.01 -0.217 0 L-8 37 M R
3 Love Ac… -0.0656 0.171 -0.347 0.216 1 L-63 22 M Other
4 Sense a… 2.23 0.616 1.21 3.24 1 L-125 30 F Other
# ℹ 4 more variables: reviews <dbl>, ebert <dbl>, romance <dbl>, action <dbl>Converting from Logit to Probability Scale
For Die Hard, our predicted logit(bechdel pass) = -0.6135, with 90% CI (-1.0104, -0.2167).
- If logit(bechdel pass) = -0.6135, then odds(bechdel pass) = exp(-0.6135), and pr(bechdel pass) = exp(-0.6135) / (1 + exp(-0.6135)) = 0.5415/1.5415 = 0.351
- If logit(bechdel pass) = -1.0104, then odds(bechdel pass) = exp(-1.0104), and pr(bechdel pass) = exp(-1.0104) / (1 + exp(-1.0104)) = 0.3641/1.3641 = 0.267
- If logit(bechdel pass) = -0.2167, then odds(bechdel pass) = exp(-0.2167), and pr(bechdel pass) = exp(-0.2167) / (1 + exp(-0.2167)) = 0.8052/1.8052 = 0.446
Predicted prob(bechdel pass) = 0.351 with 90% confidence interval (0.267, 0.446) for Die Hard using fit4.
Picking a Decision Rule for fit4
- Again, using
cutpointrto select a decision rule which maximizes “Sensitivity” + “Specificity”.
fit4_aug <- augment(fit4, type.predict = "response")
cp4 <- cutpointr(data = fit4_aug, .fitted, bechdel,
pos_class = 1, neg_class = 0,
method = maximize_metric, metric = sum_sens_spec)
cp4 |> select(direction, optimal_cutpoint, method, sum_sens_spec) |>
gt() |> tab_options(table.font.size = 24) |>
opt_stylize(style = 2, color = "pink")| direction | optimal_cutpoint | method | sum_sens_spec |
|---|---|---|---|
| >= | 0.5847669 | maximize_metric | 1.482034 |
Confusion Matrix for fit4
cm4 <- confusionMatrix(data = factor(fit4_aug$.fitted >= cp4$optimal_cutpoint),
reference = factor(fit4_aug$bechdel == 1), positive = "TRUE")
cm4Confusion Matrix and Statistics
Reference
Prediction FALSE TRUE
FALSE 76 49
TRUE 7 64
Accuracy : 0.7143
95% CI : (0.6456, 0.7764)
No Information Rate : 0.5765
P-Value [Acc > NIR] : 4.645e-05
Kappa : 0.4517
Mcnemar's Test P-Value : 4.281e-08
Sensitivity : 0.5664
Specificity : 0.9157
Pos Pred Value : 0.9014
Neg Pred Value : 0.6080
Prevalence : 0.5765
Detection Rate : 0.3265
Detection Prevalence : 0.3622
Balanced Accuracy : 0.7410
'Positive' Class : TRUE
fit4 PCP
Percentage of Correct Predictions (with 0.5 decision rule)
No severe multicollinearity? (fit4)
No Cook’s distance > 0.5? (fit4)
fit4 Partial Residual Plots
check_model for fit4 (1/4)
check_model for fit4 (2/4)
check_model for fit4 (3/4)
check_model for fit4 (4/4)
- Extra details for the last three plots…
OK: No outliers detected.
- Based on the following method and threshold: cook (0.5).
- For variable: (Whole model)OK: Simulated residuals appear as uniformly distributed (p = 0.095).Warning: About 93% of the residuals are inside the error bounds (~95% or higher would be good).Analysis of Deviance for fit4
Wald Statistics Response: bechdel
Factor Chi-Square d.f. P
gen_1 21.80 1 <.0001
reviews 7.16 1 0.0075
TOTAL 26.61 2 <.0001Note
Remember that this result shows sequential tests, and if you change the order of the predictors, the p values will change.
Validating Key Summaries (fit4)
index.orig training test optimism index.corrected n
Dxy 0.4862 0.4885 0.4845 0.0040 0.4822 300
R2 0.2853 0.2942 0.2797 0.0144 0.2709 300
Intercept 0.0000 0.0000 0.0098 -0.0098 0.0098 300
Slope 1.0000 1.0000 0.9574 0.0426 0.9574 300
Emax 0.0000 0.0000 0.0113 0.0113 0.0113 300
D 0.2335 0.2437 0.2283 0.0154 0.2181 300
U -0.0102 -0.0102 0.0239 -0.0341 0.0239 300
Q 0.2437 0.2539 0.2043 0.0496 0.1942 300
B 0.1924 0.1902 0.1952 -0.0050 0.1974 300
g 1.4011 1.7127 1.3617 0.3510 1.0501 300
gp 0.2455 0.2455 0.2389 0.0066 0.2389 300- C = 0.5 + Dxy, so validated C for
fit4= 0.5 + (0.4822/2) = 0.7411, validated Nagelkerke \(R^2\) = 0.2709, and validated Brier score B = 0.1974
Cross-Validating AUC for fit4
Model fit5
Spearman \(\rho^2\) plot from fit2
Model fit5: add two non-linear terms
fit5 <- glm(bechdel ~ gen_1 * romance + rcs(age, 4) +
mpa3 + reviews + ebert + action,
data = mov25, family = binomial(link = "logit"))
n_obs(fit5)[1] 196AUC: 80.50%# Indices of model performance
AIC | AICc | BIC | Tjur's R2 | RMSE | Sigma | Log_loss | Score_log
------------------------------------------------------------------------------
224.832 | 226.537 | 264.169 | 0.281 | 0.418 | 1.000 | 0.512 | -Inf
AIC | Score_spherical | PCP
---------------------------------
224.832 | 0.005 | 0.649Model fit5 parameters
Parameter | Odds Ratio | SE | 90% CI | z | p
--------------------------------------------------------------------------------
(Intercept) | 0.69 | 0.91 | [0.08, 5.87] | -0.28 | 0.780
gen 1 [F] | 8.21 | 5.57 | [2.94, 28.83] | 3.10 | 0.002
romance | 1.36 | 0.70 | [0.59, 3.21] | 0.60 | 0.548
rcs(age [ degree] | 0.94 | 0.08 | [0.81, 1.08] | -0.75 | 0.451
rcs(age [ degree] | 2.31 | 1.12 | [1.06, 5.26] | 1.73 | 0.083
rcs(age [ degree] | 0.14 | 0.15 | [0.02, 0.75] | -1.87 | 0.061
mpa3 [R] | 0.68 | 0.31 | [0.32, 1.45] | -0.83 | 0.406
mpa3 [Other] | 1.08 | 0.51 | [0.50, 2.37] | 0.17 | 0.864
reviews | 1.01 | 2.84e-03 | [1.01, 1.02] | 3.64 | < .001
ebert | 0.59 | 0.16 | [0.37, 0.91] | -1.93 | 0.053
action | 0.40 | 0.18 | [0.19, 0.82] | -2.06 | 0.039
gen 1 [F] × romance | 4.90e+06 | 3.97e+09 | [0.00, 1.84e+126] | 0.02 | 0.985 What is the fit5 equation?
(Intercept) gen_1F romance rcs(age, 4)age
-0.36512871 2.10491071 0.30870046 -0.06410644
rcs(age, 4)age' rcs(age, 4)age'' mpa3R mpa3Other
0.83895234 -1.94989001 -0.38266995 0.08117920
reviews ebert action gen_1F:romance
0.01025840 -0.53201822 -0.92353679 15.40442177 No one writes an equation for a cubic spline fit. We use Prediction Plots and Nomograms from fit5_lrm instead.
lrm version of fit5
Key Summaries for fit5_lrm include…
- C = 0.805, Nagelkerke \(R^2\) = 0.386, Brier score = 0.175
- See next slide for details.
fit5_lrm summaries
Logistic Regression Model
lrm(formula = bechdel ~ gen_1 * romance + rcs(age, 4) + mpa3 +
reviews + ebert + action, data = mov25, x = TRUE, y = TRUE)
Model Likelihood Discrimination Rank Discrim.
Ratio Test Indexes Indexes
Obs 196 LR chi2 66.27 R2 0.386 C 0.805
0 83 d.f. 11 R2(11,196)0.246 Dxy 0.610
1 113 Pr(> chi2) <0.0001 R2(11,143.6)0.320 gamma 0.610
max |deriv| 9e-05 Brier 0.175 tau-a 0.299
Coef S.E. Wald Z Pr(>|Z|)
Intercept -0.3651 1.3054 -0.28 0.7797
gen_1=F 2.1049 0.6792 3.10 0.0019
romance 0.3087 0.5140 0.60 0.5481
age -0.0641 0.0851 -0.75 0.4512
age' 0.8390 0.4836 1.73 0.0827
age'' -1.9499 1.0405 -1.87 0.0609
mpa3=R -0.3827 0.4603 -0.83 0.4058
mpa3=Other 0.0812 0.4733 0.17 0.8638
reviews 0.0103 0.0028 3.64 0.0003
ebert -0.5320 0.2753 -1.93 0.0533
action -0.9235 0.4481 -2.06 0.0393
gen_1=F * romance 16.1095 1899.5014 0.01 0.9932 ROC Curve Analysis for fit5
ROC curves for all 5 models
Pseudo-\(R^2\) measures (all 5 models)
r2_res12345 <- tibble(name = c("fit1", "fit2", "fit3", "fit4", "fit5"),
McFadden = c(as.numeric(r2_mcfadden(fit1)[1]),
as.numeric(r2_mcfadden(fit2)[1]),
as.numeric(r2_mcfadden(fit3)[1]),
as.numeric(r2_mcfadden(fit4)[1]),
as.numeric(r2_mcfadden(fit5)[1])),
Nagelkerke = c(as.numeric(r2_nagelkerke(fit1)),
as.numeric(r2_nagelkerke(fit2)),
as.numeric(r2_nagelkerke(fit3)),
as.numeric(r2_nagelkerke(fit4)),
as.numeric(r2_nagelkerke(fit5))),
Tjur = c(as.numeric(r2_tjur(fit1)),
as.numeric(r2_tjur(fit2)),
as.numeric(r2_tjur(fit3)),
as.numeric(r2_tjur(fit4)),
as.numeric(r2_tjur(fit5))))
r2_res12345 |> gt() |> fmt_number(decimals = 4) |>
tab_options(table.font.size = 24) |>
opt_stylize(style = 4, color = "green")Pseudo-\(R^2\) measures (all 5 models)
| name | McFadden | Nagelkerke | Tjur |
|---|---|---|---|
| fit1 | 0.1571 | 0.2590 | 0.1813 |
| fit2 | 0.2138 | 0.3398 | 0.2549 |
| fit3 | 0.2000 | 0.3206 | 0.2397 |
| fit4 | 0.1751 | 0.2853 | 0.2082 |
| fit5 | 0.2481 | 0.3856 | 0.2814 |
Comparing Model Indices from our 5 models
What are the effect sizes in fit5_lrm?
Effects Response : bechdel
Factor Low High Diff. Effect S.E. Lower 0.95
romance 0.00 1.00 1.00 0.308700 0.51402 -0.69877
Odds Ratio 0.00 1.00 1.00 1.361700 NA 0.49720
age 11.00 28.25 17.25 1.556500 0.60354 0.37359
Odds Ratio 11.00 28.25 17.25 4.742200 NA 1.45290
reviews 109.75 282.50 172.75 1.772100 0.48642 0.81877
Odds Ratio 109.75 282.50 172.75 5.883400 NA 2.26770
ebert 3.00 4.00 1.00 -0.532020 0.27532 -1.07160
Odds Ratio 3.00 4.00 1.00 0.587420 NA 0.34245
action 0.00 1.00 1.00 -0.923540 0.44805 -1.80170
Odds Ratio 0.00 1.00 1.00 0.397110 NA 0.16502
gen_1 - F:M 1.00 2.00 NA 2.104900 0.67918 0.77375
Odds Ratio 1.00 2.00 NA 8.206400 NA 2.16790
mpa3 - PG-13:Other 3.00 1.00 NA -0.081179 0.47328 -1.00880
Odds Ratio 3.00 1.00 NA 0.922030 NA 0.36466
mpa3 - R:Other 3.00 2.00 NA -0.463850 0.44262 -1.33140
Odds Ratio 3.00 2.00 NA 0.628860 NA 0.26412
Upper 0.95
1.3162000
3.7291000
2.7394000
15.4780000
2.7255000
15.2640000
0.0075955
1.0076000
-0.0453730
0.9556400
3.4361000
31.0650000
0.8464400
2.3313000
0.4036700
1.4973000
Adjusted to: gen_1=M romance=0 Plotting the Effects for fit5_lrm
Prediction Plots for fit5_lrm
Calibration Plot for fit5_lrm
n=196 Mean absolute error=0.031 Mean squared error=0.00139
0.9 Quantile of absolute error=0.061
fit5 Hosmer-Lemeshow test
Nomogram for fit5_lrm
Predictions from fit5 for love4 movies
augment(fit5, newdata = love4, type.predict = "response") |>
select(movie, .fitted, bechdel, everything()) |>
gt() |> tab_options(table.font.size = 20) |>
opt_stylize(style = 5, color = "pink")| movie | .fitted | bechdel | mov_id | age | gen_1 | mpa3 | reviews | ebert | romance | action |
|---|---|---|---|---|---|---|---|---|---|---|
| The Manchurian Candidate | 0.1523043 | 0 | L-2 | 63 | M | PG-13 | 67 | 4.0 | 0 | 0 |
| Die Hard | 0.3427204 | 0 | L-8 | 37 | M | R | 89 | 2.0 | 0 | 1 |
| Love Actually | 0.6984687 | 1 | L-63 | 22 | M | Other | 230 | 3.5 | 1 | 0 |
| Sense and Sensibility | 1.0000000 | 1 | L-125 | 30 | F | Other | 67 | 3.5 | 1 | 0 |
CIs around our predictions?
augment(fit5, newdata = love4, type.predict = "link", se_fit = TRUE) |>
mutate(ci_90_low = .fitted - 1.645 * .se.fit,
ci_90_high = .fitted + 1.645 * .se.fit) |>
select(movie, .fitted, .se.fit, ci_90_low, ci_90_high, bechdel, everything())# A tibble: 4 × 14
movie .fitted .se.fit ci_90_low ci_90_high bechdel mov_id age gen_1 mpa3
<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <dbl> <fct> <fct>
1 The Man… -1.72 0.992 -3.35 -0.0855 0 L-2 63 M PG-13
2 Die Hard -0.651 0.640 -1.70 0.402 0 L-8 37 M R
3 Love Ac… 0.840 0.623 -0.185 1.87 1 L-63 22 M Other
4 Sense a… 17.4 810. -1315. 1350. 1 L-125 30 F Other
# ℹ 4 more variables: reviews <dbl>, ebert <dbl>, romance <dbl>, action <dbl>Converting from Logit to Probability Scale
For Love Actually, our predicted logit(bechdel pass) = 0.8400, with 90% CI (-0.1852, 1.8653).
- If logit(bechdel pass) = 0.8400, then odds(bechdel pass) = exp(0.8400), and pr(bechdel pass) = exp(0.8400) / (1 + exp(0.8400)) = 2.3164/3.3164 = 0.698
- If logit(bechdel pass) = -0.1852, then odds(bechdel pass) = exp(-0.1852), and pr(bechdel pass) = exp(-0.1852) / (1 + exp(-0.1852)) = 0.8309/1.8309 = 0.454
- If logit(bechdel pass) = 1.8653, then odds(bechdel pass) = exp(1.8653), and pr(bechdel pass) = exp(1.8653) / (1 + exp(1.8653)) = 6.4578/7.4578 = 0.866
Predicted prob(bechdel pass) = 0.698 with 90% confidence interval (0.454, 0.866) for Love Actually using fit5.
Picking a Decision Rule for fit5
- Again, using
cutpointrto select a decision rule which maximizes “Sensitivity” + “Specificity”.
fit5_aug <- augment(fit5, type.predict = "response")
cp5 <- cutpointr(data = fit5_aug, .fitted, bechdel,
pos_class = 1, neg_class = 0,
method = maximize_metric, metric = sum_sens_spec)
cp5 |> select(direction, optimal_cutpoint, method, sum_sens_spec) |>
gt() |> tab_options(table.font.size = 24) |>
opt_stylize(style = 2, color = "pink")| direction | optimal_cutpoint | method | sum_sens_spec |
|---|---|---|---|
| >= | 0.5626889 | maximize_metric | 1.516686 |
Confusion Matrix for fit5
cm5 <- confusionMatrix(data = factor(fit5_aug$.fitted >= cp5$optimal_cutpoint),
reference = factor(fit5_aug$bechdel == 1), positive = "TRUE")
cm5Confusion Matrix and Statistics
Reference
Prediction FALSE TRUE
FALSE 73 41
TRUE 10 72
Accuracy : 0.7398
95% CI : (0.6725, 0.7997)
No Information Rate : 0.5765
P-Value [Acc > NIR] : 1.474e-06
Kappa : 0.4923
Mcnemar's Test P-Value : 2.659e-05
Sensitivity : 0.6372
Specificity : 0.8795
Pos Pred Value : 0.8780
Neg Pred Value : 0.6404
Prevalence : 0.5765
Detection Rate : 0.3673
Detection Prevalence : 0.4184
Balanced Accuracy : 0.7583
'Positive' Class : TRUE
fit5 PCP
Percentage of Correct Predictions (with 0.5 decision rule)
No severe multicollinearity? (fit5)
GVIF Df GVIF^(1/(2*Df))
gen_1 1.032243 1 1.015994
romance 1.121039 1 1.058792
rcs(age, 4) 2.967515 3 1.198760
mpa3 1.452883 2 1.097887
reviews 3.115263 1 1.765011
ebert 1.271118 1 1.127439
action 1.432239 1 1.196762
gen_1:romance 1.000001 1 1.000001 gen_1=F romance age age'
1.032243 1.121039 43.630373 923.944699
age'' mpa3=R mpa3=Other reviews
617.481842 1.674308 1.691158 3.115263
ebert action gen_1=F * romance
1.271118 1.432239 1.000000 No Cook’s distance > 0.5? (fit5)
fit5 Partial Residual Plots
Component + Residual plots are not available for models with interactions
check_model for fit5 (1/4)
check_model for fit5 (2/4)
check_model for fit5 (3/4)
check_model for fit5 (4/4)
- Extra details for the last three plots…
OK: No outliers detected.
- Based on the following method and threshold: cook (0.9).
- For variable: (Whole model)OK: Simulated residuals appear as uniformly distributed (p = 0.398).Warning: Probably bad model fit. Only about 79% of the residuals are inside the error bounds.Analysis of Deviance for fit5
Wald Statistics Response: bechdel
Factor Chi-Square d.f. P
gen_1 (Factor+Higher Order Factors) 9.61 2 0.0082
All Interactions 0.00 1 0.9932
romance (Factor+Higher Order Factors) 0.36 2 0.8350
All Interactions 0.00 1 0.9932
age 6.83 3 0.0776
Nonlinear 5.57 2 0.0618
mpa3 1.26 2 0.5327
reviews 13.27 1 0.0003
ebert 3.73 1 0.0533
action 4.25 1 0.0393
gen_1 * romance (Factor+Higher Order Factors) 0.00 1 0.9932
TOTAL NONLINEAR + INTERACTION 5.57 3 0.1347
TOTAL 26.50 11 0.0055Validating Key Summaries (fit5)
index.orig training test optimism index.corrected n
Dxy 0.6100 0.6602 0.5700 0.0903 0.5197 300
R2 0.3856 0.4383 0.3433 0.0950 0.2906 300
Intercept 0.0000 0.0000 -0.0097 0.0097 -0.0097 300
Slope 1.0000 1.0000 0.6926 0.3074 0.6926 300
Emax 0.0000 0.0000 0.0831 0.0831 0.0831 300
D 0.3330 0.3909 0.2901 0.1007 0.2323 300
U -0.0102 -0.0102 0.0454 -0.0556 0.0454 300
Q 0.3432 0.4011 0.2448 0.1563 0.1869 300
B 0.1749 0.1628 0.1872 -0.0244 0.1993 300
g 4.6196 5.1042 3.4482 1.6560 2.9636 300
gp 0.2990 0.3224 0.2759 0.0465 0.2525 300- C = 0.5 + Dxy, so validated C for
fit5= 0.5 + (0.5197/2) = 0.7599, validated Nagelkerke \(R^2\) = 0.2906, and validated Brier score B = 0.1993
Cross-Validating AUC for fit5
Comparing Our Five Models
compare_parameters() results (1 and 2)
Parameter | fit1 | fit2
------------------------------------------------------------
(Intercept) | 0.29 (-0.22, 0.79) | -0.05 (-1.61, 1.50)
age | -0.02 (-0.04, 0.00) | 0.02 (-0.01, 0.04)
gen 1 [F] | 2.81 ( 1.79, 3.83) | 2.59 ( 1.53, 3.64)
mpa3 [Other] | | 0.10 (-0.65, 0.85)
reviews | | 7.15e-03 ( 0.00, 0.01)
ebert | | -0.53 (-0.98, -0.08)
mpa3 [R] | | -0.30 (-1.04, 0.45)
action | | -0.77 (-1.47, -0.06)
romance | | 0.52 (-0.25, 1.29)
------------------------------------------------------------
Observations | 196 | 196compare_parameters() results (3 and 4)
Parameter | fit3 | fit4
----------------------------------------------------------------
(Intercept) | 0.53 (-0.83, 1.89) | -0.96 (-1.53, -0.39)
gen 1 [F] | 2.71 ( 1.67, 3.75) | 2.93 ( 1.90, 3.96)
reviews | 5.14e-03 ( 0.00, 0.01) | 3.89e-03 ( 0.00, 0.01)
ebert | -0.47 (-0.88, -0.05) |
action | -0.72 (-1.34, -0.09) |
----------------------------------------------------------------
Observations | 196 | 196compare_parameters() results (2 and 5)
Parameter | fit2 | fit5
-------------------------------------------------------------------------
(Intercept) | -0.05 (-1.61, 1.50) | -0.37 ( -2.51, 1.78)
gen 1 [F] | 2.59 ( 1.53, 3.64) | 2.10 ( 0.99, 3.22)
mpa3 [R] | -0.30 (-1.04, 0.45) | -0.38 ( -1.14, 0.37)
mpa3 [Other] | 0.10 (-0.65, 0.85) | 0.08 ( -0.70, 0.86)
reviews | 7.15e-03 ( 0.00, 0.01) | 0.01 ( 0.01, 0.01)
ebert | -0.53 (-0.98, -0.08) | -0.53 ( -0.98, -0.08)
romance | 0.52 (-0.25, 1.29) | 0.31 ( -0.54, 1.15)
action | -0.77 (-1.47, -0.06) | -0.92 ( -1.66, -0.19)
age | 0.02 (-0.01, 0.04) |
rcs(age [ degree] | | 0.84 ( 0.04, 1.63)
rcs(age [ degree] | | -1.95 ( -3.66, -0.24)
rcs(age [ degree] | | -0.06 ( -0.20, 0.08)
gen 1 [F] × romance | | 15.40 (-1316.65, 1347.46)
-------------------------------------------------------------------------
Observations | 196 | 196compare_performance() results
Summary Statistics
| Model | fit1 |
fit2 |
fit3 |
fit4 |
fit5 |
|---|---|---|---|---|---|
| Model df | 2 | 8 | 4 | 2 | 11 |
| C (unvalidated) | 0.7199 | 0.7921 | 0.7792 | 0.7430 | 0.8050 |
| C (validated) | 0.7183 | 0.7581 | 0.7670 | 0.7411 | 0.7599 |
| Nagelkerke \(R^2\) (un.) | 0.2590 | 0.3398 | 0.3206 | 0.2853 | 0.3856 |
| Nagelkerke \(R^2\) (val.) | 0.2469 | 0.2674 | 0.2905 | 0.2709 | 0.2906 |
| Tjur \(R^2\) (unval.) | 0.181 | 0.255 | 0.240 | 0.208 | 0.281 |
| McFadden \(R^2\) (un.) | 0.157 | 0.213 | 0.200 | 0.175 | 0.248 |
Confusion Matrix Summaries
| Model | fit1 |
fit2 |
fit3 |
fit4 |
fit5 |
|---|---|---|---|---|---|
| Cutpoint | 0.533 | 0.695 | 0.563 | 0.584 | 0.563 |
| Accuracy | 0.663 | 0.714 | 0.730 | 0.714 | 0.740 |
| Sensitivity | 0.513 | 0.531 | 0.628 | 0.566 | 0.637 |
| Specificity | 0.868 | 0.964 | 0.868 | 0.916 | 0.880 |
| Pos Pre Val | 0.841 | 0.952 | 0.866 | 0.901 | 0.878 |
| Neg Pre Val | 0.567 | 0.602 | 0.632 | 0.608 | 0.640 |
| Kappa | 0.356 | 0.458 | 0.472 | 0.452 | 0.492 |
More Model Summaries
| Model | fit1 |
fit2 |
fit3 |
fit4 |
fit5 |
|---|---|---|---|---|---|
| Mean absolute error | 0.018 | 0.026 | 0.018 | 0.040 | 0.031 |
| Mean squared error | 0.00046 | 0.00112 | 0.00052 | 0.00248 | 0.00139 |
| 0.9 quant abs error | 0.036 | 0.053 | 0.033 | 0.084 | 0.061 |
| Hosmer-Lemeshow p | 0.747 | 0.358 | 0.265 | 0.186 | 0.617 |
| Brier (unval.) | 0.1998 | 0.1811 | 0.1845 | 0.1924 | 0.1749 |
| Validated Brier | 0.2042 | 0.1989 | 0.1927 | 0.1974 | 0.1993 |
| % Correct Preds. | 0.6002 | 0.6362 | 0.6288 | 0.6134 | 0.6491 |
| Cross-Val AUC | 0.7293 | 0.7594 | 0.7433 | 0.7367 | 0.7242 |
Predictions for love4 movies
| Movie | bechdel |
fit1 |
fit2 |
fit3 |
fit4 |
fit5 |
|---|---|---|---|---|---|---|
| The Manchurian Candidate | 0 | 0.283 | 0.344 | 0.271 | 0.332 | 0.152 |
| Die Hard | 0 | 0.394 | 0.284 | 0.341 | 0.351 | 0.343 |
| Love Actually | 1 | 0.465 | 0.675 | 0.520 | 0.484 | 0.698 |
| Sense & Sensibility | 1 | 0.925 | 0.908 | 0.876 | 0.903 | 1.000 |
What’s not in this example?
- Missing data and single/multiple imputation with
miceoraregImpute() - Partitioning the data into training and test samples
- Evaluation of a logistic regression model in a test sample
- Fitting a Bayesian logistic regression model
The support1000 example covers each of these pieces, in addition to the majority of the material in this mov25 example.
Session Information
R version 4.4.2 (2024-10-31 ucrt)
Platform: x86_64-w64-mingw32/x64
Running under: Windows 11 x64 (build 22631)
Locale:
LC_COLLATE=English_United States.utf8
LC_CTYPE=English_United States.utf8
LC_MONETARY=English_United States.utf8
LC_NUMERIC=C
LC_TIME=English_United States.utf8
Package version:
abind_1.4-8 ape_5.8.1 askpass_1.2.1
backports_1.5.0 base64enc_0.1-3 bayestestR_0.15.1
bestglm_0.37.3 bigD_0.3.0 bit_4.5.0.1
bit64_4.6.0.1 bitops_1.0.9 blob_1.2.4
boot_1.3-31 broom_1.0.7 bslib_0.9.0
cachem_1.1.0 callr_3.7.6 car_3.1-3
carData_3.0-5 caret_7.0-1 cellranger_1.1.0
checkmate_2.3.2 chk_0.10.0 class_7.3-22
cli_3.6.3 clipr_0.8.0 clock_0.7.2
cluster_2.1.6 cobalt_4.5.5 coda_0.19-4.1
codetools_0.2-20 colorspace_2.1-1 commonmark_1.9.2
compiler_4.4.2 conflicted_1.2.0 correlation_0.8.6
cowplot_1.1.3 cpp11_0.5.1 crayon_1.5.3
crosstalk_1.2.1 curl_6.2.0 cutpointr_1.2.0
data.table_1.16.4 datasets_4.4.2 datawizard_1.0.0
DBI_1.2.3 dbplyr_2.5.0 Deriv_4.1.6
DHARMa_0.4.7 diagram_1.6.5 digest_0.6.37
doBy_4.6.25 doParallel_1.0.17 dplyr_1.1.4
dtplyr_1.3.1 e1071_1.7-16 easystats_0.7.3
effectsize_1.0.0 emmeans_1.10.7 estimability_1.5.1
evaluate_1.0.3 fansi_1.0.6 farver_2.1.2
fastmap_1.2.0 fontawesome_0.5.3 forcats_1.0.0
foreach_1.5.2 foreign_0.8-88 Formula_1.2-5
fs_1.6.5 future_1.34.0 future.apply_1.11.3
gap_1.6 gap.datasets_0.0.6 gargle_1.5.2
gdata_3.0.1 generics_0.1.3 ggplot2_3.5.1
ggrepel_0.9.6 glmnet_4.1-8 globals_0.16.3
glue_1.8.0 gmodels_2.19.1 googledrive_2.1.1
googlesheets4_1.1.1 gower_1.0.2 graphics_4.4.2
grDevices_4.4.2 grid_4.4.2 gridExtra_2.3
grpreg_3.5.0 gt_0.11.1 gtable_0.3.6
gtools_3.9.5 hardhat_1.4.1 haven_2.5.4
highr_0.11 Hmisc_5.2-2 hms_1.1.3
htmlTable_2.4.3 htmltools_0.5.8.1 htmlwidgets_1.6.4
httpuv_1.6.15 httr_1.4.7 ids_1.0.1
insight_1.0.1 ipred_0.9-15 isoband_0.2.7
iterators_1.0.14 janitor_2.2.1 jquerylib_0.1.4
jsonlite_1.8.9 juicyjuice_0.1.0 KernSmooth_2.23.24
knitr_1.49 labeling_0.4.3 labelled_2.14.0
later_1.4.1 lattice_0.22-6 lava_1.8.1
lazyeval_0.2.2 leaps_3.2 lifecycle_1.0.4
listenv_0.9.1 lme4_1.1-36 lmtest_0.9.40
lubridate_1.9.4 magrittr_2.0.3 markdown_1.13
MASS_7.3-64 Matrix_1.7-1 MatrixModels_0.5-3
memoise_2.0.1 methods_4.4.2 mgcv_1.9-1
microbenchmark_1.5.0 mime_0.12 minqa_1.2.8
mitools_2.4 modelbased_0.8.9 ModelMetrics_1.2.2.2
modelr_0.1.11 multcomp_1.4-28 munsell_0.5.1
mvtnorm_1.3-3 naniar_1.1.0 nlme_3.1-166
nloptr_2.1.1 nnet_7.3-20 norm_1.0.11.1
numDeriv_2016.8.1.1 openssl_2.3.2 parallel_4.4.2
parallelly_1.42.0 parameters_0.24.1 patchwork_1.3.0
pbkrtest_0.5.3 performance_0.13.0 pillar_1.10.1
pkgconfig_2.0.3 plotly_4.10.4 pls_2.8-5
plyr_1.8.9 polspline_1.1.25 prettyunits_1.2.0
pROC_1.18.5 processx_3.8.5 prodlim_2024.06.25
progress_1.2.3 progressr_0.15.1 promises_1.3.2
proxy_0.4-27 ps_1.8.1 purrr_1.0.2
qgam_1.3.4 quantreg_6.00 R6_2.5.1
ragg_1.3.3 rappdirs_0.3.3 rbibutils_2.3
RColorBrewer_1.1.3 Rcpp_1.0.14 RcppArmadillo_14.2.2.1
RcppEigen_0.3.4.0.2 Rdpack_2.6.2 reactable_0.4.4
reactR_0.6.1 readr_2.1.5 readxl_1.4.3
recipes_1.1.0 reformulas_0.4.0 rematch_2.0.0
rematch2_2.1.2 report_0.6.0 reprex_2.1.1
reshape2_1.4.4 rlang_1.1.5 rmarkdown_2.29
rms_7.0-0 rpart_4.1.24 rstudioapi_0.17.1
rvest_1.0.4 sandwich_3.1-1 sass_0.4.9
scales_1.3.0 see_0.10.0 selectr_0.4.2
shape_1.4.6.1 shiny_1.10.0 snakecase_0.11.1
sourcetools_0.1.7.1 SparseM_1.84-2 sparsevctrs_0.2.0
splines_4.4.2 SQUAREM_2021.1 stats_4.4.2
stats4_4.4.2 stringi_1.8.4 stringr_1.5.1
survey_4.4-2 survival_3.8-3 sys_3.4.3
systemfonts_1.2.1 tableone_0.13.2 textshaping_1.0.0
TH.data_1.1-3 tibble_3.2.1 tidyr_1.3.1
tidyselect_1.2.1 tidyverse_2.0.0 timechange_0.3.0
timeDate_4041.110 tinytex_0.54 tools_4.4.2
tzdb_0.4.0 UpSetR_1.4.0 utf8_1.2.4
utils_4.4.2 uuid_1.2.1 V8_6.0.1
vctrs_0.6.5 viridis_0.6.5 viridisLite_0.4.2
visdat_0.6.0 vroom_1.6.5 withr_3.0.2
xfun_0.50 xml2_1.3.6 xtable_1.8-4
yaml_2.3.10 zoo_1.8-12 
432 Class 08 | 2025-02-06 | https://thomaselove.github.io/432-2025/